Answer :
To determine which table represents a linear function, we need to analyze the relationship between [tex]\(x\)[/tex] and [tex]\(y\)[/tex] in each table. A linear function will have the same slope between any two consecutive points.
### Table Analysis:
1. First Table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 2 & 7 \\ \hline 3 & 11 \\ \hline 4 & 15 \\ \hline \end{array} \][/tex]
- Slope between points (1, 3) and (2, 7):
[tex]\[ \frac{7 - 3}{2 - 1} = \frac{4}{1} = 4 \][/tex]
- Slope between points (2, 7) and (3, 11):
[tex]\[ \frac{11 - 7}{3 - 2} = \frac{4}{1} = 4 \][/tex]
- Slope between points (3, 11) and (4, 15):
[tex]\[ \frac{15 - 11}{4 - 3} = \frac{4}{1} = 4 \][/tex]
Since the slope is consistent (4), this table represents a linear function.
2. Second Table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 2 & 8 \\ \hline 3 & 15 \\ \hline 4 & 21 \\ \hline \end{array} \][/tex]
- Slope between points (1, 3) and (2, 8):
[tex]\[ \frac{8 - 3}{2 - 1} = \frac{5}{1} = 5 \][/tex]
- Slope between points (2, 8) and (3, 15):
[tex]\[ \frac{15 - 8}{3 - 2} = \frac{7}{1} = 7 \][/tex]
- Slope between points (3, 15) and (4, 21):
[tex]\[ \frac{21 - 15}{4 - 3} = \frac{6}{1} = 6 \][/tex]
Since the slopes are not consistent, this table does not represent a linear function.
3. Third Table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 2 & 9 \\ \hline 3 & 3 \\ \hline 4 & 9 \\ \hline \end{array} \][/tex]
- Slope between points (1, 3) and (2, 9):
[tex]\[ \frac{9 - 3}{2 - 1} = \frac{6}{1} = 6 \][/tex]
- Slope between points (2, 9) and (3, 3):
[tex]\[ \frac{3 - 9}{3 - 2} = \frac{-6}{1} = -6 \][/tex]
- Slope between points (3, 3) and (4, 9):
[tex]\[ \frac{9 - 3}{4 - 3} = \frac{6}{1} = 6 \][/tex]
The slopes are not consistent, so this table does not represent a linear function.
4. Fourth Table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 2 & 9 \\ \hline 3 & 27 \\ \hline 4 & 81 \\ \hline \end{array} \][/tex]
- Slope between points (1, 3) and (2, 9):
[tex]\[ \frac{9 - 3}{2 - 1} = \frac{6}{1} = 6 \][/tex]
- Slope between points (2, 9) and (3, 27):
[tex]\[ \frac{27 - 9}{3 - 2} = \frac{18}{1} = 18 \][/tex]
- Slope between points (3, 27) and (4, 81):
[tex]\[ \frac{81 - 27}{4 - 3} = \frac{54}{1} = 54 \][/tex]
The slopes are not consistent, indicating this table does not represent a linear function.
### Conclusion:
Only the first table has a consistent slope between all pairs of points, making it the table that represents a linear function. Thus, the answer is the first table.
### Table Analysis:
1. First Table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 2 & 7 \\ \hline 3 & 11 \\ \hline 4 & 15 \\ \hline \end{array} \][/tex]
- Slope between points (1, 3) and (2, 7):
[tex]\[ \frac{7 - 3}{2 - 1} = \frac{4}{1} = 4 \][/tex]
- Slope between points (2, 7) and (3, 11):
[tex]\[ \frac{11 - 7}{3 - 2} = \frac{4}{1} = 4 \][/tex]
- Slope between points (3, 11) and (4, 15):
[tex]\[ \frac{15 - 11}{4 - 3} = \frac{4}{1} = 4 \][/tex]
Since the slope is consistent (4), this table represents a linear function.
2. Second Table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 2 & 8 \\ \hline 3 & 15 \\ \hline 4 & 21 \\ \hline \end{array} \][/tex]
- Slope between points (1, 3) and (2, 8):
[tex]\[ \frac{8 - 3}{2 - 1} = \frac{5}{1} = 5 \][/tex]
- Slope between points (2, 8) and (3, 15):
[tex]\[ \frac{15 - 8}{3 - 2} = \frac{7}{1} = 7 \][/tex]
- Slope between points (3, 15) and (4, 21):
[tex]\[ \frac{21 - 15}{4 - 3} = \frac{6}{1} = 6 \][/tex]
Since the slopes are not consistent, this table does not represent a linear function.
3. Third Table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 2 & 9 \\ \hline 3 & 3 \\ \hline 4 & 9 \\ \hline \end{array} \][/tex]
- Slope between points (1, 3) and (2, 9):
[tex]\[ \frac{9 - 3}{2 - 1} = \frac{6}{1} = 6 \][/tex]
- Slope between points (2, 9) and (3, 3):
[tex]\[ \frac{3 - 9}{3 - 2} = \frac{-6}{1} = -6 \][/tex]
- Slope between points (3, 3) and (4, 9):
[tex]\[ \frac{9 - 3}{4 - 3} = \frac{6}{1} = 6 \][/tex]
The slopes are not consistent, so this table does not represent a linear function.
4. Fourth Table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 2 & 9 \\ \hline 3 & 27 \\ \hline 4 & 81 \\ \hline \end{array} \][/tex]
- Slope between points (1, 3) and (2, 9):
[tex]\[ \frac{9 - 3}{2 - 1} = \frac{6}{1} = 6 \][/tex]
- Slope between points (2, 9) and (3, 27):
[tex]\[ \frac{27 - 9}{3 - 2} = \frac{18}{1} = 18 \][/tex]
- Slope between points (3, 27) and (4, 81):
[tex]\[ \frac{81 - 27}{4 - 3} = \frac{54}{1} = 54 \][/tex]
The slopes are not consistent, indicating this table does not represent a linear function.
### Conclusion:
Only the first table has a consistent slope between all pairs of points, making it the table that represents a linear function. Thus, the answer is the first table.
